Kernel of a characteristic action on an abelian group implies strongly image-potentially characteristic

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., kernel of a characteristic action on an abelian group) must also satisfy the second subgroup property (i.e., strongly image-potentially characteristic subgroup)
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Statement

Statement with symbols

Suppose H is a subgroup of a group G, and there exists an abelian group V, and a homomorphism \alpha:G \to \operatorname{Aut}(V) such that H equals the kernel of \alpha and V is a characteristic subgroup of the semidirect product V \rtimes G.

Then, there exists a group K with a surjective homomorphism \rho:K \to G such that both the kernel of \rho and \rho^{-1}(H) are characteristic subgroups of K.

Facts used

  1. Characteristicity is centralizer-closed
  2. Quotient group acts on abelian normal subgroup

Proof

Given: H \le G, an abelian group V, a homomorphism \alpha:G \to \operatorname{Aut}(V) with kernel H. V is characteristic in V \rtimes G.

To prove: There exists a group K with a surjective homomorphism \rho:K \to G such that both the kernel of \rho and \rho^{-1}(H) are characteristic subgroups of K.

Proof: Let K = V \rtimes G and \rho:K \to G be the quotient map.

  1. By assumption, V, the kernel of \rho is characteristic in K.
  2. C_K(V) is characteristic in K: This follows from fact (1).
  3. C_K(V) = V \times H = \rho^{-1}(H): Since V is abelian, the quotient K/V acts on V by conjugation (fact (2)). In other words, any two elements in the same coset of V act the same way on V by conjugation. Thus, a coset of V centralizes V if and only if the element of G in the coset centralizes V, which happens if and only if the element is in H. Thus, C_K(V) = V \rtimes H = V \times H (because the action is trivial) and is equal to \rho^{-1}(H).

By step (1), the kernel of \rho is characteristic in K. Combining steps (2) and (3) yields that \rho^{-1}(H) is also characteristic in K, completing the proof.